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Let $X$ be a subset of $\textbf{R}$, and let $f:X\rightarrow\textbf{R}$ be a function. Then the following two statements are logically equivalent:

(a) $f$ is uniformly continuous on $X$

(b) Whenever $(x_{n})_{n=0}^{\infty}$ and $(y_{n})_{n=0}^{\infty}$ are two equivalent sequences of elements of $X$, the sequences $(f(x_{n}))_{n=0}^{\infty}$ and $(f(y_{n}))_{n=0}^{\infty}$ are also equivalent.

MY ATTEMPT

I am mainly interested in proving the implication $(\Rightarrow)$.

Since $f$ is uniformly continuous on $X$, for every $\varepsilon > 0$, there exists a $\delta > 0$ s.t. that for every $x,y\in X$ \begin{align*} |x - y| \leq \delta \Longrightarrow |f(x) - f(y)| \leq \varepsilon \end{align*}

Moreover, since $x_{n}$ and $y_{n}$ are equivalent, for every $\delta > 0$, there exists a natural $N\geq 0$ such that \begin{align*} n\geq N \Longrightarrow |x_{n} - y_{n}| \leq \delta \end{align*}

Consequently, if we substitute $x = x_{n}$ and $y = y_{n}$, it results that \begin{align*} n\geq N \Longrightarrow |x_{n} - y_{n}| \leq \delta \Longrightarrow |f(x_{n}) - f(y_{n})| \leq \varepsilon \end{align*}

And the proposed resul holds.

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Let's say $f$ "has property U" if whenever $x_n$ and $y_n$ are equivalent sequences in the domain of $f$, $f(x_n)$ and $f(y_n)$ are also equivalent. (This is not standard terminology, I just made it up on the spot for ease of writing.)

Here are some strategies to prove the desired implication:

  1. Assume $f$ has property U and then show that $f$ is uniformly continuous. This is a direct proof.
  2. Assume $f$ is not uniformly continuous and then show that $f$ does not have property U (by constructing a counterexample). This is a proof by contraposition.
  3. Assume $f$ has property U and is not uniformly continuous and derive a contradiction to some other "unrelated" fact. This is proof by contradiction (in the strict sense).

In my experience many people write proofs by contradiction for implications that are really just slightly obfuscated proofs by contraposition, because they show $p \wedge \neg q \Rightarrow \neg p$: in other words the thing they contradict is the actual assumption of the implication, rather than an actual unrelated fact, which means they could've just done a proof by contraposition in the first place.

Meanwhile, in this particular problem, trying for a direct proof leaves you with an awkward assumption to use; you are left to to somehow use sequences to "probe" the modulus of continuity of $f$, and it's not particularly obvious how to do that.

So I will explain how to use proof by contraposition here.

Assume that $f$ is not uniformly continuous, then there exists $\varepsilon > 0$ such that for all $\delta>0$ there exist $x,y$ with $|x-y|<\delta$ and $|f(x)-f(y)|>\varepsilon$. Instantiate such a $\varepsilon$, then use the statement with $\delta=1/n$ to furnish some particular $x_n,y_n$ with $|x_n-y_n|<1/n$ but $|f(x_n)-f(y_n)|>\varepsilon$. Then $x_n$ and $y_n$ are equivalent sequences whereas $f(x_n)$ and $f(y_n)$ are not equivalent sequences.

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